An electrochemical cell, consist of the following two redox couples, M^ x+ (aq)/M(s) [E_ red ^ =+0.15 V ] and Fe^ 3+ (aq)/Fe(s) [E_ red ^ =-0.036 V ]. The cell EMF (E_ cell ) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be 10^ -2 , then the value of x is ______. (Nearest integer) [Given: M is a p-block metal and 2.303RT F =0.059 V]
A. A
Given standard reduction potentials: E^ _ M^ x+ /M = +0.15 V E^ _ Fe^ 3+ /Fe = -0.036 V Since E^ _ M^ x+ /M > E^ _ Fe^ 3+ /Fe , reduction occurs at the M^ x+ /M electrode (cathode) and oxidation occurs at the Fe^ 3+ /Fe electrode (anode). Cathode reaction: (M^ x+ + x e^ - M) 3 Anode reaction: (Fe Fe^ 3+ + 3 e^ - ) x Overall balanced cell reaction: 3 M^ x+ + x Fe 3 M + x Fe^ 3+ The number of electrons transferred in the balanced reaction is n = 3x. Standard cell potential E^ _ cell is: E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = 0.15 - (-0.036) = 0.186 V Using the Nernst equation: E_ cell = E^ _ cell - 2.303RT nF Q Substitute the given values (E_ cell = 0.2057 V, Q = 10^ -2 , and 2.303RT F = 0.059 V): 0.2057 = 0.186 - 0.059 3x (10^ -2 ) 0.2057 - 0.186 = - 0.059 3x (-2) 0.0197 = 0.118 3x 3x = 0.118 0.0197 3x 6 x = 2 Answer: 2
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